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Vector Addition

If p1, p2, ...

Question

If $p_{1}, p_{2}, p_{3}$ are respectively the lengths of perpendiculars from the vertices of a triangle $A B C$ to the opposite sides then
(i) $\frac{cos A}{p_{1}} + \frac{cos B}{p_{2}} + \frac{cos C}{p_{3}} = \frac{1}{R}$
(ii) $\frac{b p_{1}}{c} + \frac{c p_{2}}{a} + \frac{a p_{3}}{b} = \frac{a^{2} + b^{2} + c^{2}}{2 R}$ .

A

True

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B

False

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Solution

The correct option is A True
Refer to the figure.

$s e g A D ⊥ s e g B C$
$s e g B E ⊥ s e g A C$
$s e g C F ⊥ s e g A B$

Let $p_{1}$ = length of seg $A D$

$p_{2}$ = length of seg $B E$

$p_{3}$ = length of seg $C F$

1) Now, Area of $△ A B C$ = $△ = \frac{1}{2} \times b \times h$

$∴ △ = \frac{1}{2} \times p_{1} \times a$

$△ = \frac{1}{2} \times p_{2} \times b$

$△ = \frac{1}{2} \times p_{3} \times c$

Thus, $\frac{1}{2} \times p_{1} \times a = \frac{1}{2} \times p_{2} \times b = \frac{1}{2} \times p_{3} \times c = △$

$∴ p_{1} = \frac{2 △}{a}$

$p_{2} = \frac{2 △}{b}$

$p_{3} = \frac{2 △}{c}$

Thus, $\frac{c o s A}{p_{1}} + \frac{c o s B}{p_{2}} + \frac{c o s C}{p_{3}}$

$= \frac{c o s A}{(\frac{2 △}{a})} + \frac{c o s B}{(\frac{2 △}{b})} + \frac{c o s C}{(\frac{2 △}{c})}$

$= \frac{a c o s A}{2 △} + \frac{b c o s B}{2 △} + \frac{c c o s C}{2 △}$

$= \frac{a c o s A + b c o s B + c c o s C}{2 △}$

$= \frac{1}{2 △} [\frac{a s i n A c o s A}{s i n A} + \frac{b s i n B c o s B}{s i n B} + \frac{c s i n C c o s C}{s i n C}]$

$= \frac{1}{△} [(\frac{a}{2 s i n A}) s i n A c o s A + (\frac{b}{2 s i n B}) s i n B c o s B + (\frac{c}{2 s i n C}) s i n C c o s C]$

$= \frac{1}{△} [R s i n A c o s A + R s i n B c o s B + R s i n C c o s C]$
where, $R$ = radius of circumcircle of triangle.

$= \frac{R}{2 △} [2 s i n A c o s A + 2 s i n B c o s B + 2 s i n C c o s C]$

$= \frac{R}{2 △} [s i n 2 A + s i n 2 B + s i n 2 C]$

$= \frac{R}{2 △} [2 s i n (\frac{2 A + 2 B}{2}) c o s (\frac{2 A - 2 B}{2}) + s i n 2 C]$

$= \frac{R}{2 △} [2 s i n (A + B) c o s (A - B) + 2 s i n C . c o s C]$

$= \frac{R}{2 △} [2 s i n (π - C) c o s (A - B) + 2 s i n C . c o s C]$

$= \frac{R}{2 △} [2 s i n C c o s (A - B) + 2 s i n C . c o s C]$

$= \frac{R}{2 △} [2 s i n C (c o s (A - B) + c o s C)]$

$= \frac{R}{2 △} [2 s i n C (c o s (A - B) + c o s (π - (A + B)))]$

$= \frac{R}{2 △} [2 s i n C (c o s (A - B) - c o s (A + B))]$

$= \frac{R}{2 △} [2 s i n C (2 s i n A s i n B)]$

$= \frac{R}{2 △} [4 s i n A s i n B s i n C]$ (1)

Now, $△ = \frac{a b c}{4 R}$
$△ = \frac{a b c}{4 (\frac{a}{2 s i n A})}$

$∴ △ = \frac{s i n A . a b c}{2 a}$

$∴ △ = \frac{s i n A . b c}{2}$

$∴ s i n A = \frac{2 △}{b c}$

Similarly, $s i n B = \frac{2 △}{a c}$ and $s i n C = \frac{2 △}{a b}$

From equation (1),

$= \frac{R}{2 △} [4 \frac{2 △}{b c} \times \frac{2 △}{a c} \times \frac{2 △}{a b}]$

$= \frac{16 R △^{2}}{a^{2} b^{2} c^{2}}$

$= \frac{16 R △^{2}}{{(a b c)}^{2}}$

$= \frac{16 R △^{2}}{{(4 R △)}^{2}}$

$= \frac{16 R △^{2}}{16 R^{2} △^{2}}$

$= \frac{1}{R}$

2) $L H S = \frac{b p_{1}}{c} + \frac{c p_{2}}{a} + \frac{a p_{3}}{b}$

$= \frac{b (2 △)}{a c} + \frac{c (2 △)}{a b} + \frac{a (2 △)}{b c}$

$= 2 △ [\frac{b^{2} + c^{2} + a^{2}}{a b c}]$

$= 2 (\frac{a b c}{4 R}) [\frac{b^{2} + c^{2} + a^{2}}{a b c}]$

$= \frac{a^{2} + b^{2} + c^{2}}{2 R}$

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